Java Program Example - Reverse an Array */

import java.util.Scanner;

public class JavaProgram
{
public static void main(String args[])
{
float size, i, j, temp;
float arr[] = new float[50];
Scanner scan = new Scanner(System.in);

System.out.print("Enter Array Size : " );
size = scan.nextFloat();

System.out.print("Enter Array Elements : " );
for(i=0; i<size; i++)
{
arr[i] = scan.nextFloat();
}

j = i - 1; // now j will point to the last element
i = 0; // and i will point to the first element

while(i<j)
{
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i++;
j--;
}

System.out.print("Now the Reverse of Array is : \n" );
for(i=0; i<size; i++)
{
System.out.print(arr[i]+ " " );
}
}
} //please help me correct this error its displaying error incompatible type : float can't be convert to int, when I use Int it works but in this situation I am dealing with float.

System.out.print("Enter Array Size : " );
change this size = scan.nextFloat();
to size = scan.nesting();

System.out.print("Enter Array Size : " );
change this size = scan.nextFloat();
to size = scan.nextInt(); you can not have array of size 2.5.

The error start from here. arr[i] = scan.nextFloat();

float size, i, j, temp;

Make your i an integer. For when you get here:
arr[i] = scan.nextFloat();

arr[1.1] = scan.nextFloat(); would be likely. And that would be out flat wrong. Since array indices can only be integers.

Just as an addition, also make size an integer. Though, it wouldn't throw an exception, but don't make it a floating point number.

klazbaba:
float size, i, j, temp;


... also make size an integer. ...

^^^

It work now, I declare everything as integer, except this. temp which is float and this still remain float arr[] = new float[50];